Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 2}{x + 1} = \dfrac{-4x - 5}{x + 1}$
Answer: Multiply both sides by $x + 1$ $ \dfrac{x^2 - 2}{x + 1} (x + 1) = \dfrac{-4x - 5}{x + 1} (x + 1)$ $ x^2 - 2 = -4x - 5$ Subtract $-4x - 5$ from both sides: $ x^2 - 2 - (-4x - 5) = -4x - 5 - (-4x - 5)$ $ x^2 - 2 + 4x + 5 = 0$ $ x^2 + 3 + 4x = 0$ Factor the expression: $ (x + 3)(x + 1) = 0$ Therefore $x = -3$ or $x = -1$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.